--- title: "CST plate: stresses at centroid" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:cst-centroid-stresses-a2" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-centroid-stresses-a2 site: BCT Engineering --- # CST plate: stresses at centroid Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2072 Asoj, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-centroid-stresses-a2 - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question A steel plate of thickness 10 mm is being loaded in the structural system as shown in the figure below. Calculate stresses at the centroid of the plate. $E = 200$ GPa, $\nu = 0.3$. [Figure: right triangular plate, 300 mm along the top edge (supported/hatched) and 400 mm along the left vertical edge; a distributed load of 10 kN/mm$^2$ [as printed] acts on the left edge] ## Short answer For the single CST, stresses are constant, so centroid stresses equal element stresses: sigma_x=0, sigma_y=0, tau_xy=-1.333 N/mm^2, indicating pure shear. The free node displacement is u2=0.00693 mm horizontally with v2=0. ## Answer A CST has constant strain, so the stresses at the centroid equal the stresses everywhere in the plate. The plate is modelled as one CST in plane stress (units N and mm). #### Assumptions - The load "10 kN/mm$^2$" is read as a line load $w = 10$ kN/m $= 10$ N/mm along the 400 mm left edge, acting horizontally (normal to the edge, in $+x$). Total load $= 10\times400 = 4000$ N, shared equally by the end nodes of the edge. - Nodes (anticlockwise): 1 (0, 0) top-left, 2 (0, -400) bottom-left, 3 (300, 0) top-right. The top edge 1-3 is supported, so $u_1=v_1=u_3=v_3=0$. - $E=2\times10^5$ N/mm$^2$, $\nu=0.3$, $t=10$ mm. #### Step 1: Geometry $$ A=\tfrac12(300)(400)=60000\ \text{mm}^2 $$ $b_1=y_2-y_3=-400,\ b_2=y_3-y_1=0,\ b_3=y_1-y_2=400$; $c_1=x_3-x_2=300,\ c_2=x_1-x_3=-300,\ c_3=x_2-x_1=0$. $$ [B]=\frac{1}{120000}\begin{bmatrix} -400&0&0&0&400&0\\ 0&300&0&-300&0&0\\ 300&-400&-300&0&0&400\end{bmatrix} $$ #### Step 2: [D] for plane stress $$ [D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&0.3&0\\0.3&1&0\\0&0&0.35\end{bmatrix}=\begin{bmatrix} 219780 & 65934 & 0 \\ 65934 & 219780 & 0 \\ 0 & 0 & 76923 \end{bmatrix}\ \text{N/mm}^2 $$ #### Step 3: Stiffness of the free node 2 $$ [k]_{22}=tA[B]^T[D][B]\Big|_{u_2,v_2}=\begin{bmatrix} 288462 & 0 \\ 0 & 824176 \end{bmatrix}\ \text{N/mm} $$ #### Step 4: Displacement Load at node 2: $F_{2x}=4000/2=2000$ N, $F_{2y}=0$. $$ u_2=\frac{2000}{288461.5}=0.00693\ \text{mm},\qquad v_2=0 $$ #### Step 5: Stresses $$ \{\varepsilon\}=[B]\{u\}=\begin{bmatrix}0\\0\\-1.733\times10^{-5}\end{bmatrix},\qquad \{\sigma\}=[D]\{\varepsilon\}=\begin{bmatrix}0\\0\\-1.333\end{bmatrix}\ \text{N/mm}^2 $$ Check: the element force at node 2 in the $x$ direction, from $tA[B]^T\{\sigma\}$, is $\tfrac{t}{2}(b_2\sigma_x+c_2\tau_{xy})=5\times(-300)\times(-1.333)=2000$ N, which balances the applied $F_{2x}$. **Answer:** at the centroid $\sigma_x=0$, $\sigma_y=0$, $\tau_{xy}=-1.333$ N/mm$^2$, i.e. the plate is in pure shear. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the stress at the centroid the same as at any point in the plate? A. A CST has constant strain, so stress is uniform B. The plate is symmetric C. The load acts at the centroid D. Self-weight is ignored Answer: A. [B] is constant for a CST, so strain and stress are uniform in the element. (Key point: Why centroid stress equals element stress) 2. How is the 10 N/mm edge load over the 400 mm left edge applied to nodes? A. 4000 N total, 2000 N at each end node B. 10 N at node 2 C. 4000 N at node 1 D. 40 000 N at node 2 Answer: A. 10 x 400 = 4000 N, shared equally; the supported node absorbs its half, so node 2 gets 2000 N. (Key point: Load conversion) 3. Which node is free and which DOFs are solved? A. Node 2 (u2, v2), since the top edge 1-3 is supported B. Node 1 C. Node 3 D. All three nodes Answer: A. Top edge nodes 1 and 3 are fixed, leaving node 2 with u2 = 0.00693 mm and v2 = 0. (Key point: Boundary condition and free node) 4. What stress state results and how is it checked? A. Pure shear tau_xy = -1.333 N/mm^2; check 4000/(300 x 10) = 1.333 B. Uniaxial sigma_x = 1.333 N/mm^2 C. Biaxial tension of 1.333 N/mm^2 D. Pure bending Answer: A. eps_x = eps_y = 0 and only gamma_xy is nonzero, so the 4000 N is carried in shear over the 300 mm x 10 mm section. 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